Question

The displacement 's' of a particle at time 't' seconds is given by s= t^2(t-3) , then find the velocity of partical at t=2 sec​

Answer 1

Solution :-

The displacement 's' of a particle at time 't' seconds is given by

$\rm :\longmapsto\:s = {t}^{2}(t - 3)$

$\rm :\longmapsto\:s = {t}^{3} - 3 {t}^{2}$

We know that,

Velocity of a particle of displacement 's' at time 't' is given by

$\bf :\longmapsto\:v = \dfrac{ds}{dt}$

So,

Velocity is

$\rm :\longmapsto\:v = \dfrac{d}{dt} s$

$\rm :\longmapsto\:v = \dfrac{d}{dt}( {t}^{3} - {3t}^{2})$

$\rm :\longmapsto\:v = \dfrac{d}{dt} {t}^{3} - \dfrac{d}{dt} {3t}^{2}$

$\rm :\longmapsto\:v = {3t}^{2} - 6t$

$\: \: \: \: \: \: \: \: \: \: \: \: \because\boxed{\pink{\bf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\rm :\longmapsto\:v_{(2 sec)} \: = 3 \times {(2)}^{2} - 6 \times 2$

$\rm :\longmapsto\:v_{(2 sec)} = 12 - 12$

$\rm :\longmapsto\:v_{(2 sec)} = 0$

$\bf\implies \:Particle \: comes \: to \: rest.$

$\boxed{\pink{\bf\:Additional \: Information}}$

Displacement :-

• Its a vector quantity and is defined as shortest distance of the object between initial position and final position.

Velocity :-

• It is a vector quantity which is defined as the rate at which position of the object changes.

Acceleration :-

• It is a vector quantity and is defined as the rate at which Velocity of the object changes.

Answer 2

Answer:

the displacement S of a particle at time t second is given by S=t^2(t-3), then find the velocity of the particle at t=25 second